Question

Air contains $23\%$ oxygen and $77\%$ nitrogen by weight. The percentage of $O_2$ by volume is:

• A. $28.1$
• B. $20.7$
• C. $21.8$
• D. $23.0$

Answer 1

Answer: (B)

$20.7$

Let, the mixture be of 100g.

Therefore, $O_2=23g=\frac{23}{32}=0.719$ moles

$N_2=77g=\frac{77}{28}=2.75$ moles

By avogadro's law,

$\frac{V_{O_2}}{V_{N_2}}=\frac{n_{O_2}}{n_{N_2}}$

$\therefore \frac{V_{O_2}}{V_{N_2}}=\frac{0.719}{2.75}$

$\therefore V_{O_2}=0.26V_{N_2}$

The % of $O_2$ by volume= $\frac{V_{O_2}}{V_{O_2}+V_{N_2}}\times 100$

$=\frac{0.26V_{N_2}}{0.26V_{N_2}+V_{N_2}}\times 100$

$=20.7\%$

Hence, the correct option is $B$